If a ^ 2-3a + 1 = 0, find the value of 2A ^ 2-3a + A ^ - 1

If a ^ 2-3a + 1 = 0, find the value of 2A ^ 2-3a + A ^ - 1

A ^ 2-3a + 1 = 0; a = 0, the equation does not hold, so a is not equal to 0, both sides are divided by AA-3 + 1 / a = 0; a + 1 / a = 3; the two roots of quadratic equation AX ^ 2 + BX + C = 0 are x = [- B ± (b ^ 2-4ac) ^ (1 / 2)] / 2A when B ^ 2-4ac > = 0; the root of a ^ 2-3a + 1 = 0 is x = [3 ± 5 ^ (1 / 2)] / 22a ^ 2-3a + A ^ - 1

1/(a-1)-(1-3a)/(2-2a)

The original formula is 1 / (A-1) + (1-3a) / [2 (A-1)] and the simplest common denominator is 2 (A-1)
=2/[2(a-1)]+(1-3a)/[2(a-1)]
=(2+1-3a)/[2(a-1)]
=(3-3a)/[2(a-1)]
=-3 (A-1) / [2 (A-1)], about (A-1)
=-3/2

If 13A + 1 and 2A − 73 are opposite numbers, then a = ()
A. 43B. 10C. -43D. -10

From the meaning of the question: (13a + 1) + (2a − 73) = 0, remove the denominator, get a + 3 + 2a-7 = 0, transfer the term, merge 3A = 4, divide both sides of the equation by 3, get a = 43

3a-1 and 1-2a are opposite numbers, then a is equal to

They are opposite numbers, namely 3a-1 = - (1-2a) or - (3a-1) = 1-2a
If you choose 3a-1 = - (1-2a),
By moving - (1-2a) to the expression on the left, 3a-1 + 1-2a = 0 is obtained, and 5A = 0 is deduced, that is, a = 0;
If you choose - (3a-1) = 1-2a
We can also get 3a-1 + 1-2a = 0 by moving the - (3a-1) term to the right expression, and deduce 5A = 0, that is, a = 0;